以知一个三角形的两角分别是45°,60°,它们的夹边的边长是1,求最小边长我的解法:由题目知∠A=45°,∠B=60°,c=1,∠C=π-105°=75°∠C>∠B>∠A所以最小边长是对应∠A的a边由正弦定理得a=sinA*c/sinC=sin4
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![以知一个三角形的两角分别是45°,60°,它们的夹边的边长是1,求最小边长我的解法:由题目知∠A=45°,∠B=60°,c=1,∠C=π-105°=75°∠C>∠B>∠A所以最小边长是对应∠A的a边由正弦定理得a=sinA*c/sinC=sin4](/uploads/image/z/11627054-62-4.jpg?t=%E4%BB%A5%E7%9F%A5%E4%B8%80%E4%B8%AA%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E4%B8%A4%E8%A7%92%E5%88%86%E5%88%AB%E6%98%AF45%C2%B0%2C60%C2%B0%2C%E5%AE%83%E4%BB%AC%E7%9A%84%E5%A4%B9%E8%BE%B9%E7%9A%84%E8%BE%B9%E9%95%BF%E6%98%AF1%2C%E6%B1%82%E6%9C%80%E5%B0%8F%E8%BE%B9%E9%95%BF%E6%88%91%E7%9A%84%E8%A7%A3%E6%B3%95%EF%BC%9A%E7%94%B1%E9%A2%98%E7%9B%AE%E7%9F%A5%E2%88%A0A%3D45%C2%B0%2C%E2%88%A0B%3D60%C2%B0%2Cc%3D1%2C%E2%88%A0C%3D%CF%80-105%C2%B0%3D75%C2%B0%E2%88%A0C%3E%E2%88%A0B%3E%E2%88%A0A%E6%89%80%E4%BB%A5%E6%9C%80%E5%B0%8F%E8%BE%B9%E9%95%BF%E6%98%AF%E5%AF%B9%E5%BA%94%E2%88%A0A%E7%9A%84a%E8%BE%B9%E7%94%B1%E6%AD%A3%E5%BC%A6%E5%AE%9A%E7%90%86%E5%BE%97a%3DsinA%2Ac%2FsinC%3Dsin4)
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